Take the number x^y, with y an integer greater than 0. Every time you divide x^y by x, you decrease the value of the exponent by 1 because (a^b)/(a^c) = a^(b-c). Eventually, you'll reach x^1, which is x. If you divide this by x, you get 1, which is x^0.

can I call on you all too? what on earth do you do with 0^0? I mean 1 to the power of zero is certainly 1, but zero to the power of zero? My calculator tells me: domain error. lol, I need to know though, is it 1 too? I mean I could say that zero is like just getting very very close to 0, but I'm not too sure. I have a function and I have to know if it equals zero or not and it contains 0^0, and if that is 1 then the function in some cases makes more sense in others in makes not sense at all if it does equal 1. I'm confused .

You fools why are you complicating it. You have to know but the rule of dividing indices. When dividing you take each away, so 10^3/10^3 = 10^0 right, ye follow? Let us convert that to numbers. 1000/1000 = 1. Hence observe 10^0 to equal 1. boosh in yo eye.

If you are thinking in terms of limits, then this describes the problem:

Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.

They also mention this:

The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift.

I will have to search my cellar for the older copies of Schlomilch's Zeitschrift. Ha ha.

I have a function and I have to know if it equals zero or not and it contains 0^0, and if that is 1 then the function in some cases makes more sense in others in makes not sense at all if it does equal 1.

Sounds like a calculus limit problem with indeterminate forms. You use logarithms to solve the 0^0 ones.

I was trying to explain it in an informal proof using the simple properties of exponents. Examples are great, but if you want to show that it holds for all real numbers (or some other set), you need to use variables more.

Yeah sorry I'm just seriously ignorant when it comes to maths. Here at school the standard of people's maths is shockingly high! They seem to understand all this x d y √(x+z)√10^3^ with shocking ease; but you move on to any language work, english or especially other languages, and they are shockingly terrible. Shocking is an understatement. At x d y √(x+z)√10^3^ sincostan3x/^[2dfgsdfgs(b-x^7)]/2a I tend to lean back on my chair stare at a wall and then a girl and stare at the board again and laugh. I suppose I can't fathom how this would be of any use in real life. You're better off doing weights.

I think you are right, and to put it another way, 4*0^(-4) is undefined, because 4/0 is undefined.

Eureka wrote:hmmmmmm… 0^(-4):

x^(-4) as x-->0 = infinity0^x as x-->-4 = 0

I think the problem is that the expression "0^x as x-->-4" is itself invalid, because when x is near -4 it's negative, and that means we have zero in the denominator. In other words, 0^x is undefined, for all x<0.

Eureka wrote:…Still, for most purposes, a/0=infinity.

Maybe what you meant to say was, as b approaches zero, the expression a/b increases to infinity. Infinity is not a number, and the expression a/0 is not a valid expression. So we can't say that a/0=infinity. Both sides of this equation are not valid.

Democritus wrote:Maybe what you meant to say was, as b approaches zero, the expression a/b increases to infinity.

Yes, I know infinity is not a number, but I wrote “=infinity” rather than “-->infinity” for simplicity.

Democritus wrote:

Eureka wrote:hmmmmmm… 0^(-4):

x^(-4) as x-->0 = infinity0^x as x-->-4 = 0

I think the problem is that the expression "0^x as x-->-4" is itself invalid, because when x is near -4 it's negative, and that means we have zero in the denominator. In other words, 0^x is undefined, for all x<0.

I’ve consulted my old notes. If you want to evaluate a^b at a=0, b=-4:

first set b=-4
so, a^b = a^-4
evaluate the limit as a-->0:
a^-4-->infinity

then set a=0
so, a^b=0^b=0
evaluate the limit as b-->-4
0=0 regardless of the value of b

Because the methods do not agree, we have no solution. Both 0/x=0 and 0/x-->infinty are slightly correct, but technically incorrect.